solveing an equation help?

Question

So the main question I am doing is... " The area of a right triangle is 60 sq. in. The longer leg is 1 less than twice the shorter leg. Find the length of each leg. Show all work" what I have so far is "60= (1\2)(x)(2x-1)
60= (1\2)(2x^2-x)..." I cannot figure out how to solve this equation to figure out the variable so I can figure out the sides than use the pythag. theorem for the hypoteneous. Can someone help?

do not solve my whol problem just help with the equation

60= (1\2)(x)(2x-1)

Answer 1

just use the quadratic formula...

what you have is 2x^2-x-120=0
which is of the form ax^2+bx+c=0

plug it into x = (-b +/- Sqrt[b^2-4ac])/(2a) and use this value to determine the other leg length...

Answer 2

60= (1\2)(x)(2x-1)
120 = 2x^2 - x (by multiplying by 2 and distributing)
0 = 2x^2 - x - 120
The quadratic equation states:
If ax^2 + bx + c = 0
then x = [-b +/- sqrt(b^2-4ac)]/2a
In your case a = 2, b = -1, c = -120
so x = [1 +/- sqrt(1-4(2)(-120))]/(2)(2)
so x = [1 +/- sqrt(961)]/4
so x = [1 +/- 31]/4
so x = 32/4=8 or -30/4=-7.5
Since a leg of a triangle cannot be negative, x = 8 is one side, 2x-1 = 15 is the other leg, and the hypotenuse is sqrt(8^2+15^2) = 17.
Check: Area = (1/2)bh = (1/2)(8)(15) = 60 yes!

Note: the quadratic could have been factored into (x-8)(2x+15) to get the answer faster!

Answer 3

Your equation is correct for several reasons. Because you refer to the sides as "legs", this must be a right triangle even if it doesn't say so. In that case, when you use the formula for the area of a triangle, A=1/2 bh, you can use the two legs for base and height. You do not need the length of the hypotenuse and thus the Pythagorean thm. Just finish solving the equation you came up with. It does come out very nice, though isn't easy. Use the quadratic formula if you can't factor it.

You can do it!

Answer 4

Let s=length of shorter leg and
......L=length of longer leg.

From the problem, L=2s-1

Area=(s x L)/2=60

s x L=s(2s-1)=2s^2-s

=> s^2-s/2=60 or s^2-s/2-60=0

This is a quadratic equation, for which you can solve for s, which is the length of the shorter leg. After you find s, you can plug that into the formula L=2s-1 to find the length of the longer leg.

Answer 5

what you got is right
1/2(x)(2x - 1) = 60
simplifying
2x^2 - x - 120 = 0
2x^2 - 16x + 15x -120 =0
2x(x -8) + 15(x - 8) = 0
(2x + 15)(x- 8) = 0
ignoring negative value
x = 8
so small leg =x = 8 units
long leg = 2x - 1 = 2(8) -1 = 15 units
so third leg(hypotnuse) = sqrt(8^2 + 15^2)
= sqrt((64 + 225))
= sqrt(289)
= 17 units

Answer 6

Expand the equation:
2x^2 - x = 120
2x^2 - x - 120 = 0
Now solve the quadratic equation for x, discarding the negative value since length cannot be negative.

Answer 7

60=(1/2) (2x^2-x)
120=(2x^2-x)
2x^2 - x - 120 = 0
2x^2 - 16x + 15x - 120 = 0
2x^2 + 15x - 16x - 120 = 0
x(2x+15) - 8(2x+15) = 0
(2x+15)(x-8) = 0
therefore either 2x+15=0 or x-8=0
therefore x= -7.5 which is not possible in your case
hence x=8